When you graph the function (x^4)-2, the graph crosses the x axis at two points and forms one big flat trough. Since the graph is of degree 4, and has two (symmetric) roots(zeros) I would have assumed that each zero is of multiplicity 2. But I thought that a graph does not cross the x-axis at even-multiplicity zeros, but in this case, they do?

Could somebody please help me out and provide some surefire general rules for cases like this. Thanks in advance.

  • $\begingroup$ The other two roots are complex. So each root which you are talking of is of multiplicity 1. $\endgroup$ – Shailesh Oct 10 '15 at 12:33

If you factor this polynomial completely, you'll see that $$x^4 - 2 = (x^2 + \sqrt{2})(x^2 - \sqrt{2}) = (x + \sqrt[4]{2}i)(x - \sqrt[4]{2}i)(x + \sqrt[4]{2})(x - \sqrt[4]{2}).$$ Over $\mathbb{R}$, we have to stop at $$x^4 - 2 = (x^2 + \sqrt{2})(x^2 - \sqrt{2}) = (x^2 + \sqrt{2})(x + \sqrt[4]{2})(x - \sqrt[4]{2}),$$ because the first quadratic factor only factors over the complex numbers. This polynomial has only 2 real roots, each of multiplicity 1.

Over $\mathbb{C}$, every polynomial of degree $k$ has exactly $k$ roots, counting multiplicity (this is known as the fundamental theorem of algebra).


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